Performance Specifications and Compensation Design

A lead compensator $G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}$ ($\alpha<1$) adds positive phase near the new gain crossover frequency, which raises the phase margin. Its maximum phase lead $\phi_m$ occurs at $\omega_m=\dfrac{1}{T\sqrt\alpha}$, with $\sin\phi_m=\dfrac{1-\alpha}{1+\alpha}$, and there it adds a gain of $10\log(1/\alpha)$ dB. (The design below uses the Bode plot method.)

Step 1: Gain K from the error constant

$$K_a=\lim_{s\to0}s^2G(s)=\lim_{s\to0}\frac{s^2K}{s^2(1+0.25s)}=K=10$$ $$G(s)=\frac{10}{s^2(1+0.25s)}$$

Step 2: Uncompensated system

$$|G(j\omega)|=\frac{10}{\omega^2\sqrt{1+0.0625\omega^2}},\qquad \phi=-180^\circ-\tan^{-1}0.25\omega$$

Asymptotic Bode magnitude: $-40$ dB/dec through $20\log10=20$ dB at $\omega=1$; corner at $\omega=4$, then $-60$ dB/dec.

Solving $|G|=1$: $\omega_{gc}=2.85$ rad/s and

$$PM=180^\circ+(-180^\circ-\tan^{-1}0.713)=-35.5^\circ$$

The uncompensated system is unstable (phase is always below $-180^\circ$).

Step 3: Phase lead required

$$\phi_{req}=PM_d-PM_{unc}+\epsilon = 35^\circ-(-35.5^\circ)+\epsilon\approx 75^\circ\text{–}85^\circ$$

A single lead stage cannot do this: as the gain crossover moves right, the plant phase falls further (the $(1+0.25s)$ lag), and a check shows one lead section can give at most about $18.6^\circ$ of PM for this plant. So we use two identical lead sections in cascade, each giving about half the lead (a double-lead compensator):

$$G_c(s)=\left(\frac{1+Ts}{1+\alpha Ts}\right)^2$$

Step 4: Choose α

Trial 1: $\phi_m=45^\circ$ per stage ($\alpha=0.172$) gives PM ≈ 34.8° (just short). Trial 2: $\phi_m\approx50^\circ$ per stage. Take $\alpha=0.13$:

$$\sin\phi_m=\frac{1-0.13}{1+0.13}=0.770\Rightarrow\phi_m=50.3^\circ\ \text{per stage}\ (100.7^\circ\ \text{total})$$

Step 5: New gain crossover frequency

Two stages add $2\times10\log(1/\alpha)=20\log(1/0.13)=17.72$ dB at $\omega_m$. So the new crossover is where the uncompensated magnitude is $-17.72$ dB:

$$\frac{10}{\omega^2\sqrt{1+0.0625\omega^2}}=0.13\Rightarrow\omega_m=\omega_{gc}'=6.39\ \text{rad/s}$$

Step 6: Time constants

$$\begin{aligned} T&=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{6.39\times\sqrt{0.13}}=0.434\ \text{s}\\ \alpha T&=0.13\times0.434=0.0564\ \text{s} \end{aligned}$$

Corner frequencies: zero at $1/T=2.30$ rad/s, pole at $1/\alpha T=17.7$ rad/s.

$$G_c(s)=\left(\frac{1+0.434s}{1+0.0564s}\right)^2$$

The attenuation of the lead network $\alpha^2$ is made up by an amplifier of gain $1/\alpha^2=59.2$, so that the DC gain and hence $K_a=10$ are unchanged.

Step 7: Compensated system and check

$$G_c(s)G(s)=\frac{10(1+0.434s)^2}{s^2(1+0.25s)(1+0.0564s)^2}$$

At $\omega=6.39$ rad/s:

$$\begin{aligned} \phi&=-180^\circ-\tan^{-1}(1.60)+2\times50.3^\circ\\ &=-180^\circ-58.0^\circ+100.7^\circ=-137.3^\circ\\ PM&=180^\circ-137.3^\circ=42.7^\circ\ (\ge35^\circ) \end{aligned}$$

GM of the compensated system ≈ 12.9 dB (phase crossover at 17.1 rad/s). All closed-loop poles are in the LHP ($-27.0$, $-4.08\pm j7.90$, $-2.14\pm j1.14$).

R(s)->( )->[Amp 59.2]->[(s+2.3)/(s+17.7)]^2 ->[G(s)]-+->C
       ^-                                            |
       |_____________________________________________|

Answer: $G_c(s)=\left(\dfrac{1+0.434s}{1+0.0564s}\right)^2$ (two lead stages, α = 0.13 each) with K = 10 gives Ka = 10 s⁻² and PM ≈ 43° (> 35°).

True. A derivative (D) controller is never used alone; it is always combined with a proportional and/or integral controller, as PD or PID.

Justification

A derivative controller gives an output proportional to the rate of change of the error:

$$u(t)=K_d\frac{de(t)}{dt},\qquad G_c(s)=K_ds$$

1. No action on constant error: if the error is constant (even large), $de/dt=0$, so the controller output is zero. A steady error is never corrected, so D alone cannot hold the output at the set point.
2. Blocks DC signals: $G_c(0)=0$. In the loop it acts like a zero at the origin, which can cancel the plant's integrator and reduce the system type, so the steady-state error grows.
3. Noise amplification: $|G_c(j\omega)|=K_d\omega$ rises with frequency, so high-frequency noise is amplified strongly. With a P term the useful low-frequency action dominates.
4. Anticipatory, not corrective: D action predicts the error trend and adds damping; it only improves the transient part of the response.

How it is used

With proportional action (PD):

$$G_c(s)=K_p+K_ds=K_p(1+T_ds)$$

• The P part removes and holds the error.
• The D part adds a zero, increases damping, reduces overshoot and settling time, and improves stability.

Example: in a position servo, a PD controller lets the motor slow down as it nears the target (D) while still pushing it there (P). A pure D controller would stop acting once the motor is at rest away from the target.

A lead compensator $G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}$ ($\alpha<1$) is designed on the Bode plot: it supplies its maximum lead $\phi_m$ ($\sin\phi_m=\frac{1-\alpha}{1+\alpha}$) at the new gain crossover $\omega_m=\frac{1}{T\sqrt\alpha}$, where it adds $10\log(1/\alpha)$ dB of gain.

Step 1: K from the velocity error constant

$$K_v=\lim_{s\to0}sG(s)=K=1000\ \text{s}^{-1}$$ $$G(s)=\frac{1000}{s(1+0.1s)(1+0.001s)}$$

Step 2: Uncompensated Bode plot

Corners at 10 and 1000 rad/s. Asymptotic magnitude: 60 dB at $\omega=1$, $-20$ dB/dec to 10 (40 dB), $-40$ dB/dec to 1000 ($-40$ dB), then $-60$ dB/dec.

$$\phi(\omega)=-90^\circ-\tan^{-1}0.1\omega-\tan^{-1}0.001\omega$$

From $|G|=1$: $\omega_{gc}=99.5$ rad/s, $\phi=-179.94^\circ$, so

$$PM_{unc}\approx0^\circ$$

(and $\omega_{pc}=100$ rad/s, GM ≈ 0 dB). The system is on the verge of instability.

Step 3: Required phase lead

$$\phi_m=PM_d-PM_{unc}+\epsilon=45^\circ-0^\circ+10^\circ=55^\circ$$

A margin of $\epsilon=10^\circ$ is taken because the crossover will move to a higher frequency where the plant phase is lower (a 5° margin gives only 44° and fails the check).

Step 4: α

$$\alpha=\frac{1-\sin55^\circ}{1+\sin55^\circ}=\frac{1-0.819}{1+0.819}=0.0994\approx0.1$$

With $\alpha=0.1$: $\phi_m=\sin^{-1}(0.9/1.1)=54.9^\circ$.

Step 5: New gain crossover frequency

The lead adds $10\log(1/0.1)=10$ dB at $\omega_m$, so place $\omega_m$ where $|G(j\omega)|=-10$ dB:

$$\frac{1000}{\omega\sqrt{1+0.01\omega^2}\sqrt{1+10^{-6}\omega^2}}=0.316\Rightarrow\omega_m\approx176\ \text{rad/s}$$

Step 6: Time constants

$$\begin{aligned} T&=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{176\times0.316}=0.018\ \text{s}\\ \alpha T&=0.0018\ \text{s} \end{aligned}$$

Zero at $1/T=55.6$ rad/s, pole at $1/\alpha T=556$ rad/s.

$$G_c(s)=\frac{1+0.018s}{1+0.0018s}$$

An amplifier of gain $1/\alpha=10$ makes up the lead network's attenuation, so $K_v$ stays 1000.

Step 7: Compensated system and check

$$G_c(s)G(s)=\frac{1000(1+0.018s)}{s(1+0.1s)(1+0.001s)(1+0.0018s)}$$

At $\omega_{gc}'=176.8$ rad/s:

$$\begin{aligned} \phi&=-90^\circ-\tan^{-1}17.68-\tan^{-1}0.177+\tan^{-1}3.18-\tan^{-1}0.318\\ &=-90^\circ-86.8^\circ-10.0^\circ+72.5^\circ-17.6^\circ=-131.9^\circ\\ PM&=180^\circ-131.9^\circ=48.1^\circ\ (\ge45^\circ) \end{aligned}$$

GM ≈ 17.5 dB at $\omega_{pc}\approx696$ rad/s.

 dB 60 |\  uncompensated  -20
    37 |  \__ 10    compensated
     0 |------\----x----------- w
       |    100 \  176.8
       |         \__  -40/-60
 lead adds phase between 55.6 and 556 rad/s

Answer: $G_c(s)=\dfrac{1+0.018s}{1+0.0018s}$ (α = 0.1, T = 0.018 s) with K = 1000 gives Kv = 1000 s⁻¹ and PM ≈ 48° (≥ 45°).

Derivative (rate or tachometer) feedback control feeds back a signal proportional to the rate of change of the output, $K_t\,dc/dt$, through an inner (minor) loop, in addition to the usual unity feedback. In a position servo this signal comes from a tachogenerator on the motor shaft.

Block diagram

R(s)    E(s)        +----------------+     C(s)
--->(+)---->(+)---->| wn^2/(s(s+2z wn))|--+---->
     ^-      ^-     +----------------+  |
     |       |                          |
     |       +-------[ Kt s ]<----------+
     |                                  |
     +----------------------------------+

Transfer function

Inner loop (forward $G(s)=\dfrac{\omega_n^2}{s(s+2\zeta\omega_n)}$, feedback $K_ts$):

$$\begin{aligned} G_1(s)&=\frac{G}{1+GK_ts}=\frac{\omega_n^2}{s(s+2\zeta\omega_n)+K_t\omega_n^2s}=\frac{\omega_n^2}{s\left(s+2\zeta\omega_n+K_t\omega_n^2\right)} \end{aligned}$$

Closing the unity outer loop:

$$\frac{C(s)}{R(s)}=\frac{\omega_n^2}{s^2+\left(2\zeta\omega_n+K_t\omega_n^2\right)s+\omega_n^2}$$

Comparing with $s^2+2\zeta'\omega_ns+\omega_n^2$:

$$\zeta'=\zeta+\frac{K_t\omega_n}{2},\qquad \omega_n'=\omega_n$$

Effect on transient performance

• Damping increases ($\zeta'>\zeta$) while $\omega_n$ is unchanged.
• Peak overshoot reduces, since $M_p=e^{-\pi\zeta'/\sqrt{1-\zeta'^2}}$ falls as $\zeta'$ rises.
• Settling time reduces: $t_s=4/(\zeta'\omega_n)$.
• Rise time increases slightly (slower initial rise).
• Unlike derivative error (PD) control, it adds no zero to the closed loop, so it does not cause the extra overshoot a zero can produce.

Drawback: steady-state error

$$K_v=\lim_{s\to0}sG_1(s)=\frac{\omega_n}{2\zeta+K_t\omega_n},\qquad e_{ss}(\text{ramp})=\frac{2\zeta+K_t\omega_n}{\omega_n}$$

So the ramp error increases; it is usually offset by raising the forward gain.

Example: $\omega_n=4$ rad/s, $\zeta=0.25$ ($M_p=44\%$). With $K_t=0.15$: $\zeta'=0.25+0.15\times4/2=0.55$, and $M_p$ falls to about 12.6%.

Lead compensator: $G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}$, $\alpha<1$, maximum phase lead $\phi_m$ with $\sin\phi_m=\frac{1-\alpha}{1+\alpha}$ at $\omega_m=\frac{1}{T\sqrt\alpha}$, where it adds $10\log(1/\alpha)$ dB.

Step 1: K from $K_v$

$$K_v=\lim_{s\to0}sG(s)=K=10\ \text{s}^{-1}\Rightarrow G(s)=\frac{10}{s(1+0.2s)}$$

Step 2: Uncompensated system

Asymptotic magnitude: 20 dB at $\omega=1$ with $-20$ dB/dec, corner at $1/0.2=5$ rad/s, then $-40$ dB/dec. Phase $\phi=-90^\circ-\tan^{-1}0.2\omega$.

Gain crossover: $\dfrac{10}{\omega\sqrt{1+0.04\omega^2}}=1\Rightarrow0.04\omega^4+\omega^2-100=0$

$$\begin{aligned} \omega^2&=\frac{-1+\sqrt{1+16}}{0.08}=39.04\Rightarrow\omega_{gc}=6.25\ \text{rad/s}\\ PM&=180^\circ-90^\circ-\tan^{-1}(1.25)=90^\circ-51.3^\circ=38.7^\circ \end{aligned}$$

Step 3: Required phase lead

$$\phi_m=50^\circ-38.7^\circ+5^\circ=16.3^\circ$$

(5° added because the crossover moves right, where the plant has more lag.)

Step 4: α

$$\alpha=\frac{1-\sin16.3^\circ}{1+\sin16.3^\circ}=\frac{1-0.281}{1+0.281}=0.561$$

Step 5: New gain crossover

Lead gain at $\omega_m$: $10\log(1/0.561)=2.51$ dB. Find where $|G|=-2.51$ dB:

$$\frac{10}{\omega\sqrt{1+0.04\omega^2}}=\sqrt{0.561}=0.749\Rightarrow\omega_m=7.44\ \text{rad/s}$$

Step 6: T

$$\begin{aligned} T&=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{7.44\times0.749}=0.179\approx0.18\ \text{s}\\ \alpha T&=0.561\times0.179=0.10\ \text{s} \end{aligned}$$ $$G_c(s)=\frac{1+0.18s}{1+0.1s}$$

Zero at 5.56 rad/s, pole at 10 rad/s. An amplifier of gain $1/\alpha\approx1.8$ makes up the network's attenuation, keeping $K_v=10$.

Step 7: Compensated system

$$G_c(s)G(s)=\frac{10(1+0.18s)}{s(1+0.2s)(1+0.1s)}$$

Check at the new crossover $\omega_{gc}'=7.47$ rad/s:

$$\begin{aligned} \phi&=-90^\circ+\tan^{-1}1.345-\tan^{-1}1.494-\tan^{-1}0.747\\ &=-90^\circ+53.4^\circ-56.2^\circ-36.8^\circ=-129.6^\circ\\ PM&=180^\circ-129.6^\circ=50.4^\circ \end{aligned}$$

GM = ∞ (phase never reaches −180°).

Bode diagram of the compensated system

Asymptotes: $-20$ dB/dec from 20 dB at $\omega=1$; corner 5 (pole) → $-40$; corner 5.56 (zero) → $-20$; corner 10 (pole) → $-40$ dB/dec.

 dB 40 |\  -20 dB/dec
    20 |  \  (w=1)
       |    \     compensated
     0 |------\---x------------- w
       |     5 \ 7.47 -20 between
       |   5.56 \__   5.56 and 10
       |          \__ -40 after 10
 deg -90|__
     -130|    \___o  PM = 50.4 deg
     -180|-------------\_______

Answer: $G_c(s)=\dfrac{1+0.18s}{1+0.1s}$ (α ≈ 0.56) with K = 10 gives Kv = 10 s⁻¹ and PM ≈ 50° at ω ≈ 7.5 rad/s.

A PD (proportional–derivative) controller produces a control signal proportional to the error plus the rate of change of the error:

$$u(t)=K_pe(t)+K_d\frac{de(t)}{dt},\qquad G_c(s)=K_p+K_ds=K_p(1+T_ds),\ T_d=\frac{K_d}{K_p}$$

          +-->[ Kp ]-----+
 e(t) ----|              (+)---> u(t)
          +-->[ Kd d/dt ]+

Principle

• The P part acts on the present error.
• The D part acts on the trend of the error. When the error is falling fast (output rushing toward the set point), $de/dt$ is negative and the D term reduces the drive before the output overshoots. So D gives anticipatory action and extra damping.
• It adds a zero at $s=-K_p/K_d$ to the open-loop transfer function, i.e. phase lead.

Working performance (second-order system)

For $G(s)=\dfrac{\omega_n^2}{s(s+2\zeta\omega_n)}$ with PD control (taking $K_p=1$):

$$\frac{C(s)}{R(s)}=\frac{\omega_n^2(1+T_ds)}{s^2+(2\zeta\omega_n+\omega_n^2T_d)s+\omega_n^2}$$ $$\zeta'=\zeta+\frac{\omega_nT_d}{2}$$

Example: with $\omega_n=2$ rad/s and $\zeta=0.25$, choosing $T_d=0.3$ s gives $\zeta'=0.25+0.3=0.55$, cutting the step overshoot from about 44% to about 16% (the added zero keeps it a little above the 12.6% of a plain ζ = 0.55 system).

A lag compensator $G_c(s)=K_c\dfrac{1+Ts}{1+\beta Ts}$ ($\beta>1$) is used here: it gives high gain at low frequency (for $K_v$) while its attenuation of $20\log\beta$ at high frequency lowers the gain crossover to a frequency where the plant already has enough phase margin.

Step 1: Gain for the required $K_v$

$$G(s)H(s)=\frac{2}{s(s+1)(s+2)}=\frac{1}{s(1+s)(1+0.5s)}\Rightarrow K_v=\lim_{s\to0}sGH=1$$

To get $K_v=5$ the compensator must give a DC gain $K_c=5$:

$$G_1(s)=\frac{5}{s(1+s)(1+0.5s)}$$

Step 2: Uncompensated (gain-adjusted) system

$$|G_1|=\frac{5}{\omega\sqrt{1+\omega^2}\sqrt{1+0.25\omega^2}},\qquad\phi=-90^\circ-\tan^{-1}\omega-\tan^{-1}0.5\omega$$

• $\omega_{pc}=\sqrt2=1.414$ rad/s, GM $=-4.4$ dB; $\omega_{gc}=1.80$ rad/s, PM $=-13^\circ$: unstable.

Step 3: New gain crossover frequency

Required phase at $\omega_{gc}'$: $-180^\circ+40^\circ+\epsilon$ with $\epsilon=10^\circ$ (for the lag of the compensator), i.e. $-130^\circ$:

$$-90^\circ-\tan^{-1}\omega-\tan^{-1}0.5\omega=-130^\circ\Rightarrow\omega_{gc}'=0.49\ \text{rad/s}$$

Step 4: β

At 0.49 rad/s, $|G_1|=18.95$ dB $=8.86$. The lag network must cut this to 0 dB:

$$20\log\beta=18.95\ \text{dB}\Rightarrow\beta=8.86$$

Step 5: T

Place the zero one decade below the new crossover so the compensator lag there is small:

$$\frac1T=\frac{\omega_{gc}'}{10}\approx0.05\ \text{rad/s}\Rightarrow T=20\ \text{s},\quad\beta T=177\ \text{s}$$

Pole at $1/\beta T=0.00565$ rad/s.

$$G_c(s)=5\,\frac{1+20s}{1+177s}=0.565\,\frac{s+0.05}{s+0.00565}$$

Step 6: Check

$$G_c(s)G(s)H(s)=\frac{5(1+20s)}{s(1+s)(1+0.5s)(1+177s)}$$

 dB 40 |\__ uncompensated
       |   \___
     0 |---x---\------------ w
       | 0.49   \ 1.8
       | compensated curve is
       | 19 dB lower above 0.05

All specifications are met (PM ≥ 40°, GM ≥ 10 dB, $K_v=5$). The price is a lower bandwidth (slower response).

Answer: $G_c(s)=5\,\dfrac{1+20s}{1+177s}$ (β ≈ 8.9, T = 20 s): Kv = 5 s⁻¹, PM ≈ 45°, GM ≈ 14 dB.

A phase-lag network $G_c(s)=\dfrac{1+Ts}{1+\beta Ts}$ ($\beta>1$) has unity DC gain and attenuates high frequencies by $20\log\beta$ dB. It lowers the gain crossover frequency to where the plant phase gives the required PM, while $K_v$ is kept by the gain $K$.

Step 1: K from $K_v$

$$K_v=\lim_{s\to0}sG(s)=K=30\Rightarrow G(s)=\frac{30}{s(1+0.1s)(1+0.2s)}$$

Step 2: Uncompensated system

Corners 5 and 10 rad/s. Asymptotes: $-20$ dB/dec (29.5 dB at $\omega=1$) to 5, $-40$ to 10, $-60$ after.

$$\phi(\omega)=-90^\circ-\tan^{-1}0.1\omega-\tan^{-1}0.2\omega$$

$\omega_{gc}=9.77$ rad/s, PM $=-17.2^\circ$; $\omega_{pc}=7.07$ rad/s, GM $=-6$ dB. The system is unstable.

Step 3: New gain crossover

Required phase: $-180^\circ+40^\circ+10^\circ=-130^\circ$ ($10^\circ$ allowance for the lag network's own phase lag).

$$\tan^{-1}0.1\omega+\tan^{-1}0.2\omega=40^\circ\Rightarrow\omega_{gc}'=2.46\ \text{rad/s}$$

Step 4: β

At 2.46 rad/s, $|G|=20.53$ dB $=10.63$:

$$20\log\beta=20.53\Rightarrow\beta\approx10.6$$

Step 5: T

Zero about one decade below $\omega_{gc}'$:

$$\frac1T\approx\frac{2.46}{10}\approx0.25\ \text{rad/s}\Rightarrow T=4\ \text{s},\quad\beta T=42.5\ \text{s}$$

Pole at $1/\beta T=0.0235$ rad/s.

$$G_c(s)=\frac{1+4s}{1+42.5s}$$

Network realisation (passive R-C lag): $T=R_2C$, $\beta=\frac{R_1+R_2}{R_2}=10.6$; e.g. $C=10\ \mu$F, $R_2=400\ \text{k}\Omega$, $R_1=3.84\ \text{M}\Omega$.

Step 6: Compensated system

$$G_c(s)G(s)=\frac{30(1+4s)}{s(1+0.1s)(1+0.2s)(1+42.5s)}$$ $$PM=180^\circ-135.4^\circ=44.6^\circ\ (\ge40^\circ),\qquad GM=13.9\ \text{dB}$$

Answer: $G_c(s)=\dfrac{1+4s}{1+42.5s}$ (β ≈ 10.6, T = 4 s) with K = 30 gives Kv = 30 s⁻¹ and PM ≈ 44.6°.

Lead network: $G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}$, $\alpha<1$; maximum lead $\sin\phi_m=\frac{1-\alpha}{1+\alpha}$ at $\omega_m=\frac{1}{T\sqrt\alpha}$, where it adds $10\log(1/\alpha)$ dB (after the $1/\alpha$ attenuation is made up by an amplifier).

Step 1: K

$$K_v=\lim_{s\to0}sG(s)=K=20\ \text{s}^{-1}\Rightarrow G(s)=\frac{20}{s(1+0.2s)}$$

Step 2: Uncompensated system

Asymptotes: $-20$ dB/dec through 26 dB at $\omega=1$; corner at 5 rad/s; then $-40$ dB/dec.

$$\frac{20}{\omega\sqrt{1+0.04\omega^2}}=1\Rightarrow0.04\omega^4+\omega^2-400=0\Rightarrow\omega^2=88.3,\ \omega_{gc}=9.40\ \text{rad/s}$$ $$PM=90^\circ-\tan^{-1}(0.2\times9.40)=90^\circ-62.0^\circ=28.0^\circ$$

Step 3: Phase lead needed

$$\phi_m=44^\circ-28.0^\circ+5^\circ=21.0^\circ$$

Step 4: α

$$\alpha=\frac{1-\sin21^\circ}{1+\sin21^\circ}=\frac{1-0.358}{1+0.358}=0.473$$

Step 5: New gain crossover

The network adds $10\log(1/0.473)=3.25$ dB at $\omega_m$. Find where $|G|=-3.25$ dB ($=\sqrt\alpha=0.688$):

$$\frac{20}{\omega\sqrt{1+0.04\omega^2}}=0.688\Rightarrow\omega_m=11.55\ \text{rad/s}$$

Step 6: T

$$T=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{11.55\times0.688}=0.126\ \text{s},\qquad\alpha T=0.0595\approx0.06\ \text{s}$$

Zero at $1/T=7.94$ rad/s, pole at $1/\alpha T=16.8$ rad/s.

$$G_c(s)=\frac{1+0.126s}{1+0.06s}$$

Passive R-C network: $\dfrac{E_o}{E_i}=\alpha\dfrac{1+Ts}{1+\alpha Ts}$ with $T=R_1C$, $\alpha=\dfrac{R_2}{R_1+R_2}$; an amplifier of gain $1/\alpha\approx2.1$ restores $K_v$. E.g. $C=1\ \mu$F, $R_1=126\ \text{k}\Omega$, $R_2=113\ \text{k}\Omega$.

Step 7: Check

$$G_c(s)G(s)=\frac{20(1+0.126s)}{s(1+0.2s)(1+0.06s)}$$ $$PM=180^\circ-135.8^\circ=44.2^\circ\approx44^\circ$$

GM = ∞ (phase stays above $-180^\circ$).

Answer: $G_c(s)=\dfrac{1+0.126s}{1+0.06s}$ (α = 0.473) with K = 20 gives Kv = 20 s⁻¹ and PM ≈ 44°.

A PID controller combines proportional, integral and derivative actions on the error. It is the most widely used controller in industry (process control of temperature, pressure, flow, level; motor speed and position control) because it can fix both transient and steady-state behaviour with three tunable gains.

$$u(t)=K_pe(t)+K_i\int_0^te\,dt+K_d\frac{de}{dt},\qquad G_c(s)=K_p+\frac{K_i}{s}+K_ds=\frac{K_ds^2+K_ps+K_i}{s}$$

        +-->[  Kp  ]---+
 e ---->+-->[ Ki/s ]--(+)--> u ---> Plant ---> c
        +-->[ Kd s ]---+

PID adds one pole at the origin and two zeros to the open loop.

Role of each term

Uses / advantages

1. Zero steady-state error to step inputs (and ramp for a type-1 plant) due to the integral term.
2. Good transient response: derivative action limits overshoot introduced by P and I.
3. Better relative stability (PD part acts like a lead compensator, PI part like a lag; PID behaves like a lag-lead compensator).
4. Simple tuning without an exact model, e.g. Ziegler–Nichols rules: $K_p=0.6K_{cr}$, $T_i=0.5P_{cr}$, $T_d=0.125P_{cr}$.
5. Available as standard industrial hardware and in PLC/DCS software.

Limitations

• D action amplifies noise, so a filtered derivative is used.
• Integral wind-up when the actuator saturates (anti-wind-up needed).

Example: in a boiler temperature loop, P gives quick heating, I removes the final offset from the set point, and D stops the temperature overshooting when the setpoint is changed.

A PI (proportional–integral) controller produces an output proportional to the error plus the integral of the error:

$$u(t)=K_pe(t)+K_i\int_0^te(t)\,dt,\qquad G_c(s)=K_p+\frac{K_i}{s}=\frac{K_p\left(s+\frac{K_i}{K_p}\right)}{s}$$

It adds a pole at the origin and a zero at $s=-K_i/K_p$ to the open-loop transfer function.

          +-->[ Kp ]----+
 e(t) ----|             (+)--> u(t)
          +-->[Ki/s]----+

Effect on time response

Consider a plant $G(s)=\dfrac{\omega_n^2}{s(s+2\zeta\omega_n)}$ with PI control. The open loop becomes

$$G_c(s)G(s)=\frac{\omega_n^2(K_ps+K_i)}{s^2(s+2\zeta\omega_n)}$$

1. System type increases by one (type 1 to type 2). Step error stays zero and the ramp steady-state error becomes zero ($K_v=\infty$). This is the main benefit: steady-state accuracy improves.
2. Order increases (2nd to 3rd order). The characteristic equation $s^3+2\zeta\omega_ns^2+K_p\omega_n^2s+K_i\omega_n^2=0$ is stable only if $2\zeta\omega_nK_p>K_i$; too large $K_i$ causes instability.
3. Transient response: the integrator adds phase lag, so damping decreases. Overshoot increases and settling time becomes longer; rise time usually decreases a little.
4. The PI controller acts as a low-pass filter (like a lag compensator), so it reduces high-frequency noise.

Example: a speed-control loop with only P control settles with a small offset; adding the integral term removes the offset, but the speed may overshoot more before settling.

To reduce (or eliminate) steady-state error, an integral-type controller, usually a PI controller (or PID), is recommended. In frequency-domain design, the equivalent is a lag compensator.

Why integral action

The steady-state error of a unity feedback system depends on the system type and error constants:

$$e_{ss}(\text{step})=\frac{1}{1+K_p},\qquad e_{ss}(\text{ramp})=\frac{1}{K_v}$$

A PI controller

$$G_c(s)=K_p+\frac{K_i}{s}=\frac{K_p(s+K_i/K_p)}{s}$$

adds a pole at the origin, so the system type increases by one:

Physically, the integral keeps growing while any error remains, so the controller output keeps changing until the error is exactly zero.

The zero at $-K_i/K_p$ is placed close to the origin so that the extra phase lag (and the effect on transient response and stability) stays small.

Alternatives

• Increasing proportional gain reduces the error but cannot remove it, and makes the response oscillatory.
• Lag compensator $\frac{1+Ts}{1+\beta Ts}$ raises the low-frequency gain by $\beta$ ($K_v$ multiplied by β) without changing the transient much.

Example: $G(s)=\frac{10}{s+2}$ with P control has step error $\frac{1}{1+5}=0.167$. With PI control the open loop has a pole at the origin and the step error becomes zero.